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\title{Work3: 3.6.1 Theoretical questions}

\author{邵盛栋 \\ 信息与计算科学 3200103951}

\begin{document}
	\maketitle
	\section*{习题解答}
	\begin{enumerate}[I]
		\item 由 $s \in \mathbb{S}_3^2$:
		$
		s(x)= \begin{cases}p(x) & \text { if } x \in[0,1] \\ (2-x)^3 & \text { if } x \in[1,2]\end{cases}
		$且$ s(0)=0, p\in \mathbb{P}_{3} $可知：
		\[ p(0)=0\quad p(1)=1\quad p'(1)=-3\quad p''(1)=6 \]
		可构造如下table of divided difference:
		\begin{table*}[ht]
			\centering
			\begin{tabular}{c|cccc}
				0&0&&&\\
				1&1&1&&\\
				1&1&-3&-4&\\
				1&1&-3&3&7
			\end{tabular}
		\end{table*}
	
		其中， $ f[x_{1},x_{1}]=\dfrac{p'(x_{1})}{1!}=-3, f[x_{1},x_{1},x_{1}]=\dfrac{p''(x_{1})}{2!}=3 $.因此， \[p(x)=0+x-4x(x-1)+7x(x-1)^{2}=7x^{3}-18x^{2}+12x\] 
		若$ s(x) $为自然三次样条，则需满足边界条件$ s''(0)=0 $, $ s''(2)=0 $.\\
		由于$ s''(2)=0 $,而$ s''(0)=-36\neq0 $,因此$ s(x) $不是自然三次样条.
		\item 	\begin{enumerate}[(a)]
			\item 由定义可知：\[ \mathbb{S}_{2}^{1}=\lbrace s:s\in C[a,b];\forall i\in[1,n-1],s|_{[x_{i},x_{i+1}]}\in\mathbb{P}_{2}\rbrace \]
			由定理3.14可知$ \mathbb{S}_{2}^{1} $是一个$ n+1 $维的线性空间，则对于给定的$ f_{i}=f(x_{i}),i=1,2,\ldots,n $,还需要一个条件来确定唯一的$ s $.
			\item 已知$ p_{i}(x_{i})=f_{i} $, $ p_{i}(x_{i+1})=f_{i+1} $, $ p_{i}'(x_{i})=m_{i} $,可构造如下table of divided difference:
			\newpage
			\begin{table*}[ht]
				\centering
				\begin{tabular}{c|ccc}
					$ x_{i} $&$ f_{i} $&&\\
					$ x_{i} $&$ f_{i} $&$ m_{i} $&\\
					$ x_{i+1} $&$ f_{i+1} $&$ K_{i} $&$ \frac{K_{i}-m_{i}}{x_{i+1}-x_{i}} $
				\end{tabular}
			\end{table*}
			
			其中, $ K_{i}=f[x_{i},x_{i+1}]=\dfrac{f_{i+1}-f_{i}}{x_{i+1}-x_{i}} $,则可得到：
			\[ p_{i}=f_{i}+m_{i}(x-x_{i})+\dfrac{\frac{f_{i+1}-f_{i}}{x_{i+1}-x_{i}}-m_{i}}{x_{i+1}-x_{i}}(x-x_{i})^{2} \]
			\item 对$ p_{i} $关于x求导可得：
			\[ p'_{i}=m_{i}+2\dfrac{\frac{f_{i+1}-f_{i}}{x_{i+1}-x_{i}}-m_{i}}{x_{i+1}-x_{i}}(x-x_{i}) \]
			则, $ p'_{i}(x_{i+1})=m_{i}+2(\dfrac{f_{i+1}-f_{i}}{x_{i+1}-x_{i}}-m_{i}) $,即$ m_{i+1}=2\dfrac{f_{i+1}-f_{i}}{x_{i+1}-x_{i}}-m_{i} $.\\
			已知$ m_{1}=f'(a) $，则由上关系式可依次计算$ m_{2},m_{3},\ldots,m_{n-1} $.
		\end{enumerate}
		\item 已知$
		s(x)= \begin{cases}s_{1}(x) & \text { if } x \in[-1,0] \\ s_{2}(x) & \text { if } x \in[0,1]\end{cases}
		$且$ s_{1}(x)=1+c(x+1)^{3} $.\\
		由于$ s(x) $是自然三次样条，则：
		\[ s_{2}(0)=s_{1}(0)=c+1\quad s'_{2}(0)=s'_{1}(0)=3c\quad s''_{2}(0)=s''_{1}(0)=6c\quad s_{2}(1)=-1\]
		可构造如下table of divided difference:
		\begin{table*}[ht]
			\centering
			\begin{tabular}{c|cccc}
				0&c+1&&&\\
				0&c+1&3c&&\\
				0&c+1&3c&3c&\\
				1&-1&-2-c&-2-4c&-2-7c
			\end{tabular}
		\end{table*}
		
		则, $ s_{2}(x)=(c+1)+3cx+3cx^{2}-(7c+2)x^{3} $,又由$ s_{2}''(1)=0 $知$ 6c-(42c+12)=0 $\\
		解得：$ c=-\dfrac{1}{3} $
		\item \begin{enumerate}[(a)]
			\item 令$
			s(x)= \begin{cases}s_{1}(x) & \text { if } x \in[-1,0] \\ s_{2}(x) & \text { if } x \in[0,1]\end{cases}
			$，对于$ s_{2}(x) $,其满足：
			\[ s_{2}(x)=f_{2}+s'_{2}(0)x+\dfrac{M_{2}}{2}x^{2}+\dfrac{s'''_{2}(0)}{6}x^{3} \]
			已知$ M_{1}=M_{3}=0,f_{2}=f(0)=1,f(1)=f(-1)=0 $,则\\
			$ f[-1,0]=1,f[0,1]=-1,f[-1,0,1]=-1 $,因此$ 2M_{2}=6f[-1,0,1]=-6 $\\
			解得$ M_{2}=-3 $,
			\[ s'_{2}(0)=f[0,1]-\dfrac{M_{2}}{3}=0\quad s'''_{2}(0)=\dfrac{M_{3}-M_{2}}{1-0}=3 \]
			代入$ s_{2}(x) $得, $ s_{2}(x)=1-\dfrac{3}{2}x^{2}+\dfrac{1}{2}x^{3} $.\\
			同理可得, $ s_{1}(x)=1-\dfrac{3}{2}x^{2}-\dfrac{1}{2}x^{3}  $.\\
			所以自然三次样条为
			\[ s(x)= \begin{cases}1-\frac{3}{2}x^{2}-\frac{1}{2}x^{3} & \text { if } x \in[-1,0] \\ 1-\frac{3}{2}x^{2}+\frac{1}{2}x^{3} & \text { if } x \in[0,1]\end{cases} \]
			\item \begin{enumerate}[(i)]
				\item 设$ g(x)=ax^{2}+bx+c $,其满足$ g(-1)=0 $, $ g(0)=1 $, $ g(1)=0 $,则$ g(x)=-x^{2}+1 $.\\
				则$ \int_{-1}^{1}\left[g^{\prime \prime}(x)\right]^2\mathrm{~d} x=\int_{-1}^{1}4 \mathrm{~d} x=8 $.\\
				而$ \int_{-1}^{1}\left[s^{\prime \prime}(x)\right]^2\mathrm{~d} x=\int_{-1}^{0}(-3x-3)^2\mathrm{~d} x+\int_{0}^{1}(3x-3)^2\mathrm{~d} x=6 $.\\
				因此，$ \int_{-1}^{1}\left[s^{\prime \prime}(x)\right]^2\mathrm{~d} x\leq\int_{-1}^{1}\left[g^{\prime \prime}(x)\right]^2\mathrm{~d} x $.
				\item 若$ g(x)=f(x) $,则：
				\[ \int_{-1}^{1}\left[g^{\prime \prime}(x)\right]^2\mathrm{~d} x=\int_{-1}^{1}\left[f^{\prime \prime}(x)\right]^2\mathrm{~d} x=\int_{-1}^{1}\dfrac{\pi^{4}(\cos\pi x+1)}{32}\mathrm{~d} x=\dfrac{\pi^{4}}{16}>6 \]
				因此，$ \int_{-1}^{1}\left[s^{\prime \prime}(x)\right]^2\mathrm{~d} x\leq\int_{-1}^{1}\left[g^{\prime \prime}(x)\right]^2\mathrm{~d} x $.
			\end{enumerate}
		综上验证，定理3.10(Mininum bending energy)成立.
		\end{enumerate}
	\item \begin{enumerate}[(a)]
			\item 由定义3.23知
			\[ B_i^{n+1}(x)=\frac{x-t_{i-1}}{t_{i+n}-t_{i-1}}B_i^n(x)+\frac{t_{i+n+1}-x}{t_{i+n+1}-t_i} B_{i+1}^n(x) \]
			令$ n=1 $,则可得$ B_{i}^{2}(x) $的表达式：
			\[ B_i^{2}(x)=\frac{x-t_{i-1}}{t_{i+1}-t_{i-1}}B_i^1(x)+\frac{t_{i+2}-x}{t_{i+2}-t_i} B_{i+1}^1(x) \]
			由例3.24结论, $ B_{i}^{1}=\hat{B_{i}} $,又由于
			\[ \hat{B}_i(x)= \begin{cases}\frac{x-t_{i-1}}{t_i-t_{i-1}} & x \in\left(t_{i-1}, t_i\right] \\ \frac{t_{i+1}-x}{t_{i+1}-t_i} & x \in\left(t_i, t_{i+1}\right] \\ 0 & \text { otherwise. }\end{cases} \]
			代入第二个式子中即可得：
			\[ B_{i}^{2}(x)=\begin{cases}\frac{\left(x-t_{i-1}\right)^2}{\left(t_{i+1}-t_{i-1}\right)\left(t_i-t_{i-1}\right)}, & x \in\left(t_{i-1}, t_i\right] ; \\ \frac{\left(x-t_{i-1}\right)\left(t_{i+1}-x\right)}{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+1}-t_i\right)}+\frac{\left(t_{i+2}-x\right)\left(x-t_i\right)}{\left(t_{i+2}-t_i\right)\left(t_{i+1}-t_i\right)}, & x \in\left(t_i, t_{i+1}\right] ; \\ \frac{\left(t_{i+2}-x\right)^2}{\left(t_{i+2}-t_i\right)\left(t_{i+2}-t_{i+1}\right)}, & x \in\left(t_{i+1}, t_{i+2}\right] ; \\ 0, & \text { otherwise. }\end{cases} \]
			\item \[ \frac{d}{dx}B_{i}^{2}(x)=\begin{cases}\frac{2\left(x-t_{i-1}\right)}{\left(t_{i+1}-t_{i-1}\right)\left(t_i-t_{i-1}\right)}, & x \in\left(t_{i-1}, t_i\right] ; \\ \frac{-2x+(t_{i+1}-t_{i-1})}{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+1}-t_i\right)}+\frac{-2x+(t_{i+2}-t_{i})}{\left(t_{i+2}-t_i\right)\left(t_{i+1}-t_i\right)}, & x \in\left(t_i, t_{i+1}\right] ; \\ \frac{-2\left(t_{i+2}-x\right)}{\left(t_{i+2}-t_i\right)\left(t_{i+2}-t_{i+1}\right)}, & x \in\left(t_{i+1}, t_{i+2}\right] ; \\ 0, & \text { otherwise. }\end{cases} \]
			则：
			\begin{equation*}
			\begin{split}
				\lim\limits_{x\to t_{i}^{-}}\frac{d}{dx}B_{i}^{2}(x)&=\frac{2\left(t_{i}-t_{i-1}\right)}{\left(t_{i+1}-t_{i-1}\right)\left(t_i-t_{i-1}\right)}=\dfrac{2}{t_{i+1}-t_{i-1}}\\
				\lim\limits_{x\to t_{i}^{+}}\frac{d}{dx}B_{i}^{2}(x)&=\frac{-2t_{i}+(t_{i+1}-t_{i-1})}{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+1}-t_i\right)}+\frac{-2t_{i}+(t_{i+2}-t_{i})}{\left(t_{i+2}-t_i\right)\left(t_{i+1}-t_i\right)}=\dfrac{2}{t_{i+1}-t_{i-1}}\\
				\lim\limits_{x\to t_{i+1}^{-}}\frac{d}{dx}B_{i}^{2}(x)&=\frac{-2t_{i+1}+(t_{i+1}-t_{i-1})}{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+1}-t_i\right)}+\frac{-2t_{i+1}+(t_{i+2}-t_{i})}{\left(t_{i+2}-t_i\right)\left(t_{i+1}-t_i\right)}=\dfrac{2}{t_{i}-t_{i+2}}\\
				\lim\limits_{x\to t_{i+1}^{+}}\frac{d}{dx}B_{i}^{2}(x)&=\frac{-2\left(t_{i+2}-t_{i+1}\right)}{\left(t_{i+2}-t_i\right)\left(t_{i+2}-t_{i+1}\right)}=\dfrac{2}{t_{i}-t_{i+2}}
			\end{split}
			\end{equation*}
			可知, $ \lim\limits_{x\to t_{i}^{-}}\frac{d}{dx}B_{i}^{2}(x)=\lim\limits_{x\to t_{i}^{+}}\frac{d}{dx}B_{i}^{2}(x), \lim\limits_{x\to t_{i+1}^{-}}\frac{d}{dx}B_{i}^{2}(x)=\lim\limits_{x\to t_{i+1}^{+}}\frac{d}{dx}B_{i}^{2}(x) $\\
			因此，$ \frac{d}{dx}B_{i}^{2}(x) $在$ t_{i} $和$ t_{i+1} $上连续.
			\item 由(b)中得到的$ \frac{d}{dx}B_{i}^{2}(x) $的表达式可知，\\
			当$ x^{*}\in(t_{i-1},t_{i}] $时，$ \frac{d}{dx}B_{i}^{2}(x)>0 $,则$ x^{*}\in(t_{i},t_{i+1} )$.\\
			$ \frac{d}{dx}B_{i}^{2}(x) $在$ (t_{i},t_{i+1}] $上的表达式可化简为:
			\[ \frac{d}{dx}B_{i}^{2}(x)=\dfrac{-2(t_{i+2}+t_{i+1}-t_{i}-t_{i-1})x}{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+1}-t_i\right)\left(t_{i+2}-t_i\right)}+\dfrac{2}{t_{i+1}-t_{i}} \]
			则$ \frac{d}{dx}B_{i}^{2}(x) $在$ (t_{i},t_{i+1}] $上是线性方程，由于\[\frac{d}{dx}B_{i}^{2}(t_{i})=\dfrac{2}{t_{i+1}-t_{i-1}}>0,\frac{d}{dx}B_{i}^{2}(t_{i+1})=\dfrac{2}{t_{i}-t_{i+2}}<0\] 
			所以在$ (t_{i},t_{i+1}] $有且仅有一个$ x^{*} $使得$ \frac{d}{dx}B_{i}^{2}(x^{*})=0 $，解得：$ x^{*}=\dfrac{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+2}-t_i\right)}{t_{i+2}+t_{i+1}-t_{i}-t_{i-1}} $
			\item 由(c)中所求$ \frac{d}{dx}B_{i}^{2}(x) $可知，$ B_{i}^{2}(x) $在$ (t_{i-1},\dfrac{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+2}-t_i\right)}{t_{i+2}+t_{i+1}-t_{i}-t_{i-1}}] $上单调递增，\\
			在$ (\dfrac{(t_{i+1}-t_{i-1})(t_{i+2}-t_i)}{t_{i+2}+t_{i+1}-t_{i}-t_{i-1}},t_{i+2}] $上单调递减.\\
			因此, 
			\[ \begin{split}
				\min B_{i}^{2}(x)&=\min(B_{i}^{2}(t_{i-1}),B_{i}^{2}(t_{i+2}))=0 \\  \max B_{i}^{2}(x)&=B_{i}^{2}(\dfrac{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+2}-t_i\right)}{t_{i+2}+t_{i+1}-t_{i}-t_{i-1}})=\dfrac{t_{i+2}-t_{i-1}}{t_{i+2}+t_{i+1}-t_{i}-t_{i-1}}<1
			\end{split} \]
			所以$ B_{i}^{2}(x)\in[0,1) $.
			\item 将$ t_{i}=i $代入原表达式得：
			\[ B_{i}^{2}(x)=\begin{cases}\frac{1}{2}(x-i+1)^{2}, & x \in\left(i-1, i\right] ; \\ -x^{2}+(2i+1)x-i^{2}-i+\frac{1}{2}, & x \in\left(i, i+1\right] ; \\ \frac{1}{2}(i+2-x)^{2}, & x \in\left(i+1, i+2\right] ; \\ 0, & \text { otherwise. }\end{cases} \]
			以$ i=1 $为例，在matlab中作图得
			\begin{figure}[ht]
				\centering
				\includegraphics[width=0.4\linewidth]{picture}
			\end{figure}
		\end{enumerate}
		\item  \[(t_{i+2}-t_{i-1})[t_{i-1},t_{i},t_{i+1},t_{i+2}](t-x)^{2}_{+}=[t_{i},t_{i+1},t_{i+2}](t-x)^{2}_{+}-[t_{i-1},t_{i},t_{i+1}](t-x)^{2}_{+}\] 
		由定理3.30得：\[ \begin{split}
			[t_{i},t_{i+1},t_{i+2}](t-x)^{2}_{+}&=(t_{i}-x)[t_{i},t_{i+1},t_{i+2}](t-x)^{1}_{+}+[t_{i+1},t_{i+2}](t-x)^{1}_{+}\\
			[t_{i-1},t_{i},t_{i+1}](t-x)^{2}_{+}&=(t_{i-1}-x)[t_{i-1},t_{i},t_{i+1}](t-x)^{1}_{+}+[t_{i},t_{i+1}](t-x)^{1}_{+}
		\end{split} \]
		所以，\[ \begin{split}
			&(t_{i+2}-t_{i-1})[t_{i-1},t_{i},t_{i+1},t_{i+2}](t-x)^{2}_{+}\\=&(t_{i+2}-x)[t_{i},t_{i+1},t_{i+2}](t-x)^{1}_{+}-(t_{i-1}-x)[t_{i-1},t_{i},t_{i+1}](t-x)^{1}_{+}\\=&(t_{i+2}-x)[\frac{(t_{i+2}-x)_{+}^{1}-(t_{i+1}-x)_{+}^{1}}{(t_{i+2}-t_{i+1})(t_{i+2}-t_{i})}-\frac{(t_{i+1}-x)_{+}^{1}-(t_{i}-x)_{+}^{1}}{(t_{i+1}-t_{i})(t_{i+2}-t_{i})}]\\+&(x-t_{i-1})[\frac{(t_{i+1}-x)_{+}^{1}-(t_{i}-x)_{+}^{1}}{(t_{i+1}-t_{i})(t_{i+1}-t_{i-1})}-\frac{(t_{i}-x)_{+}^{1}-(t_{i-1}-x)_{+}^{1}}{(t_{i}-t_{i-1})(t_{i+1}-t_{i-1})}]
		\end{split} \]
	分别将$ x\in(t_{i-1},t_{i}],x\in(t_{i},t_{i+1}],x\in(t_{i+1},t_{i+2}] $代入得：
	\[(t_{i+2}-t_{i-1})[t_{i-1},t_{i},t_{i+1},t_{i+2}](t-x)^{2}_{+}=\begin{cases}\frac{\left(x-t_{i-1}\right)^2}{\left(t_{i+1}-t_{i-1}\right)\left(t_i-t_{i-1}\right)}, & x \in\left(t_{i-1}, t_i\right] ; \\ \frac{\left(x-t_{i-1}\right)\left(t_{i+1}-x\right)}{\left(t_{i+1}-t_{i-1}\right)\left(t_{i+1}-t_i\right)}+\frac{\left(t_{i+2}-x\right)\left(x-t_i\right)}{\left(t_{i+2}-t_i\right)\left(t_{i+1}-t_i\right)}, & x \in\left(t_i, t_{i+1}\right] ; \\ \frac{\left(t_{i+2}-x\right)^2}{\left(t_{i+2}-t_i\right)\left(t_{i+2}-t_{i+1}\right)}, & x \in\left(t_{i+1}, t_{i+2}\right] ; \\ 0, & \text { otherwise. }\end{cases}\]
	因此，$ (t_{i+2}-t_{i-1})[t_{i-1},t_{i},t_{i+1},t_{i+2}](t-x)^{2}_{+}=B_{i}^{2} $
	\item 由定理3.34，对$ n\geq1 $, $ \forall x\in\mathbb{R} $有
	\[ \frac{\mathrm{d}}{\mathrm{d} x} B_i^{n+1}(x)=\frac{(n+1) B_i^{n}(x)}{t_{i+n}-t_{i-1}}-\frac{(n+1) B_{i+1}^{n}(x)}{t_{i+n+1}-t_i} . \]
	对等式两边在$ [t_{i-1},t_{i+n+1}] $上积分得
	\[ \dfrac{1}{n+1}\int_{t_{i-1}}^{t_{i+n+1}}\frac{\mathrm{d}}{\mathrm{d} x} B_i^{n+1}(x)dx=\dfrac{1}{t_{i+n}-t_{i-1}}\int_{t_{i-1}}^{t_{i+n+1}}B_i^{n}(x)dx-\dfrac{1}{t_{i+n+1}-t_i}\int_{t_{i-1}}^{t_{i+n+1}}B_{i+1}^{n}(x)dx \]
	即：
	\[ \dfrac{1}{n+1}(B_i^{n+1}(t_{i+n+1})-B_i^{n+1}(t_{i-1}))=\dfrac{1}{t_{i+n}-t_{i-1}}\int_{t_{i-1}}^{t_{i+n+1}}B_i^{n}(x)dx-\dfrac{1}{t_{i+n+1}-t_i}\int_{t_{i-1}}^{t_{i+n+1}}B_{i+1}^{n}(x)dx \]
	由于$ B_i^{n+1}(t_{i+n+1})-B_i^{n+1}(t_{i-1})=0 $,且由引理3.27知$ B_i^{n} $的支集区间为$ [t_{i-1},t_{i+n}] $, $ B_{i+1}^{n} $的支集区间为$ [t_{i},t_{i+n+1}] $.\\则$ \int_{t_{i-1}}^{t_{i+n+1}}B_i^{n}(x)dx=\int_{t_{i-1}}^{t_{i+n}}B_i^{n}(x)dx $, $ \int_{t_{i-1}}^{t_{i+n+1}}B_{i+1}^{n}(x)dx=\int_{t_{i}}^{t_{i+n+1}}B_{i+1}^{n}(x)dx $.\\
	因此，\[ \dfrac{1}{t_{i+n}-t_{i-1}}\int_{t_{i-1}}^{t_{i+n}}B_i^{n}(x)dx=\dfrac{1}{t_{i+n+1}-t_i}\int_{t_{i}}^{t_{i+n+1}}B_{i+1}^{n}(x)dx \]
	所以B样条$ B_{i}^{n}(x) $在其支集区间上的缩放积分与$ i $的取值无关.
	\item \begin{enumerate}[(a)]
		\item 当$ m=4,n=2 $时, $ \tau_{2}(x_{i},x_{i+1},x_{i+2})=[x_{i},x_{i+1},x_{i+2}]x^{4} $.\\
		由定义3.38, $ \tau_{2}(x_{i},x_{i+1},x_{i+2})=x_{i}^{2}+x_{i+1}^{2}+x_{i+2}^{2}+x_{i}x_{i+1}+x_{i}x_{i+1}+x_{i+1}x_{x+2} $.
		对于右式，构造table of divided difference：
		\begin{table*}[ht]
			\centering
			\begin{tabular}{c|ccc}
				$ x_{i} $&$ x_{i}^{4} $&&\\
				$ x_{i+1} $&$ x_{i+1}^{4} $&$ (x_{i+1}^{2}+x_{i}^{2})(x_{i+1}+x_{i}) $&\\
				$ x_{i+2} $&$ x_{i+2}^{4} $&$ (x_{i+2}^{2}+x_{i+1}^{2})(x_{i+2}+x_{i+1}) $&$ [x_{i},x_{i+1},x_{i+2}]x^{4} $
			\end{tabular}
		\end{table*}
	
		则
		\[ \begin{split}
			[x_{i},x_{i+1},x_{i+2}]x^{4}&=\dfrac{(x_{i+2}^{2}+x_{i+1}^{2})(x_{i+2}+x_{i+1})-(x_{i+1}^{2}+x_{i}^{2})(x_{i+1}+x_{i})}{x_{i+2}-x_{i}}\\
			&=\dfrac{(x_{i+2}^{3}-x_{i}^{3})+x_{i+1}^{2}(x_{i+2}-x_{i})+x_{i+1}(x_{i+2}-x_{i})^{2}}{x_{i+2}-x_{i}}\\
			&=x_{i}^{2}+x_{i+1}^{2}+x_{i+2}^{2}+x_{i}x_{i+1}+x_{i}x_{i+1}+x_{i+1}x_{x+2}
		\end{split} \]
		所以, $ \tau_{2}(x_{i},x_{i+1},x_{i+2})=[x_{i},x_{i+1},x_{i+2}]x^{4} $成立.
		\item \begin{proof}
			\[ \forall m \in \mathbb{N}^{+}, \forall i \in \mathbb{N}, \forall n=0,1, \ldots, m\]\[\tau_{m-n}\left(x_i, \ldots, x_{i+n}\right)=\left[x_i, \ldots, x_{i+n}\right] x^m \]
			由引理3.45，
			\[ \begin{aligned}
				&\left(x_{n+1}-x_1\right) \tau_k\left(x_1, \ldots, x_n, x_{n+1}\right) \\
				=& \tau_{k+1}\left(x_1, \ldots, x_n, x_{n+1}\right)-\tau_{k+1}\left(x_1, \ldots, x_n\right)-x_1 \tau_k\left(x_1, \ldots, x_n, x_{n+1}\right) \\
				=& \tau_{k+1}\left(x_2, \ldots, x_n, x_{n+1}\right)+x_1 \tau_k\left(x_1, \ldots, x_n, x_{n+1}\right)-\tau_{k+1}\left(x_1, \ldots, x_n\right)-x_1 \tau_k\left(x_1, \ldots, x_n, x_{n+1}\right) \\
				=& \tau_{k+1}\left(x_2, \ldots, x_n, x_{n+1}\right)-\tau_{k+1}\left(x_1, \ldots, x_n\right)
			\end{aligned} \]
			利用数学归纳法，当$ n=0 $，$ \tau_{m}(x_{i})=[x_{i}]x^{m} $显然成立.\\
			假设对所有$ n<m $, $ \tau_{m-n}\left(x_i, \ldots, x_{i+n}\right)=\left[x_i, \ldots, x_{i+n}\right] x^m $均成立,则
			\[ \begin{aligned}
				& \tau_{m-n-1}\left(x_i, \ldots, x_{i+n+1}\right) \\
				=& \frac{\tau_{m-n}\left(x_{i+1}, \ldots, x_{i+n+1}\right)-\tau_{m-n}\left(x_i, \ldots, x_{i+n}\right)}{x_{i+n+1}-x_i} \\
				=& \frac{\left[x_{i+1}, \ldots, x_{i+n+1}\right] x^m-\left[x_i, \ldots, x_{i+n}\right] x^m}{x_{i+n+1}-x_i} \\
				=& {\left[x_i, \ldots, x_{i+n+1}\right] x^m }
			\end{aligned} \]
                        结论成立.
		\end{proof}
	\end{enumerate}
	\end{enumerate}
	
\end{document}
